A cost and pipeline trade-off in a transportation problem

Yugoslav Journal on Operations Research
23(2013) Number 2, 197–211
DOI: 10.2298/YJOR130214030S

A COST AND PIPELINE TRADE-OFF IN A
TRANSPORTATION PROBLEM
Vikas SHARMA
Department of Mathematics, Centre for Advanced Study in Mathematics, Panjab
University, Chandigarh, India


Rita MALHOTRA
Kamla Nehru College, University of Delhi, Khel Gaon Marg, New Delhi-110049,
India


Vanita VERMA
Department of Mathematics, Centre for Advanced Study in Mathematics, Panjab
University, Chandigarh, India

Received: January, 2013 / Accepted: March, 2013
Abstract: The present paper deals with a trade off between cost and pipeline at
a given time in a transportation problem. The time lag between commissioning a
project and the time when the last consignment of goods reaches the project site
is an important factor. This motivates the study of a bi-criteria transportation
problem at a pivotal time T . An exhaustive set E of all independent cost-pipeline
pairs (called efficient pairs) at time T is constructed in such a way that each pair
corresponds to a basic feasible solution and in turn, gives an optimal transportation
schedule. A convergent algorithm has been proposed to determine non-dominated
cost pipeline pairs in a criteria space instead of scanning the decision space, where
the number of such pairs is large as compared to those found in the criteria space.

197


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Keywords:
Transportation problem, Combinatorial optimization, Bottleneck
transportation problem, Bi-criteria transportation problem, Efficient points
MSC: 90B06, 90C05, 90C08

1. INTRODUCTION
The cost minimization transportation problem is defined as:
min

cij xij

(P1 )

i∈I j∈J

subject to the following constraints:

xij = ai , ai > 0, i ∈ I, 




j∈J


xij = bj , bj > 0, j ∈ J, 


i∈I



xij ≥ 0, ∀ (i, j) ∈ I × J

(1.1)

where I is the index set of supply points, J is the index set of destinations, xij is
the amount of the product transported from ith supply point to j th destination, cij
is the per unit of cost of transportation on (i, j)th route, ai is the availability of the
product at ith supply point and bj is the requirement of the same at j th destination.
Here the aim is to minimize the cost of transporting goods totalling
ai =
bj .
i∈I

j∈J

A time minimization transportation problem (TMTP), which is a special case of
bottleneck linear programming problem, has been studied by Arora et al. [2, 3],
Bhatia et al. [5], Garfinkel et al. [7], Hammer [9], Prakash [13] , Sharma et al.
[17]and Szwarc [18]. In (TMTP), the transportation of goods from sources to
destinations is done in parallel, and its prime aim is to supply to destinations
with the required quantity within a shortest possible time. Mathematically, this
problem can be formulated as follows:

min {max tij (xij ) | xij > 0},

(1.2)

S = {X = {xij } | X satisfies (1.1)}

(1.3)

X∈S

where
and tij is the time of transportation from ith supply point to j th destination.
In the recent past, the transportation problem with more than one objective has
been solved by many researchers like Bhatia et al. [4], Glickman and Berger [8],
Khurana et al. [10], Malhotra and Puri [11], Purushotam et al. [16], Prakash
[12, 14, 15]. Bhatia et al. [4] developed an enumerative technique to obtain successive time-cost commodity in pipeline trade-off relationship in a transportation


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199

problem. Prakash [12] has solved the transportation problem with two objectives
after having accorded first and second priorities to the minimization of total cost
and duration of transportation, respectively. Purushotam et al. [16] dealt with
this problem in reverse order of priorities. In the present paper, we have discuss
cost-pipeline trade off at a pivotal time T .
Cost-pipeline tradeoff relationship in a transportation problem is relevant
in a project planning, which requires transportation of raw materials/machinary
etc. to the site of the project before it starts functioning. Besides, taking care

of transportation costs, the goods reaching on the last day have also to be taken
into consideration because the commissioning of the project is influenced in the
sense that some time is consumed even after the last consignment of goods reaches
the site, as some formalities are required to be completed before processing them
onto the machine. This time lag between the arrival of the last consignment of
goods and the time of initiation of the project indirectly means cost to the decision
maker. As in some situations, early initiation of project is desired, which is possible when the quantity of goods reaching just before the initiation of the project
is very small, but this in turn means high cost of transportation. Therefore the
problem of interest is trade- off between total cost of transportation and pipeline
at a time T (time of transportation) such that if early initiation of the project is
desired, i.e. at some time T ∗ (< T ), means higher cost of transportation, such a
time (T ) is referred to as pivotal time. The proposed algorithm determines all the
independent, non-dominated cost-pipeline pairs called efficient pairs, which correspond to basic feasible solutions (BFS) starting from the minimum cost solution
at a pivotal time T chosen. The determination of extreme points of the nondominated set in objective space in preference to the decision space is justified;
as asserted by Aneja and Nair [1], the number of extreme points of the feasible
set in objective space is in general lesser than the that in the decision space. The
proposed algorithm finds all such pairs in criteria space. Process terminates when
no more new efficient extreme points are available.
This paper is organized as follows : Definitions and notations are given in
Section 2, theoretical results have been proved in Section 3. Section 4 discusses the
procedure to solve the problem and the paper concludes with a numerical example
discussed in Section-5.

2. MATHEMATICAL FORMULATION OF THE
PROBLEM
For any X ∈ S, let T = max{tij | xij > 0}.
At any time T , define following cost minimization transportation problem, whose
optimal solution yields minimum transportation cost at the given time T .
min

X∈S

cij xij
i∈I j∈J

(P2 )


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where
cij
∞

cij =

if tij ≤ T,
if tij > T

Let the optimal value of the problem (P2 ) be denoted by Z.
Pivotal time. Time T is called pivotal time if for any other time of transportation
T ∗ (say), T ∗ > T =⇒ Z ∗ < Z and T ∗ < T =⇒ Z ∗ > Z, where Z and Z ∗ are
minimum transportation costs given by (P2 ) at times T and T ∗ , respectively.
Remark 2.1. T is a pivotal time of transportation if in each optimal basic feasible
solution (OBFS) of the problem (P2 ) there exists a cell (i, j) with tij = T, xij > 0.
Pipeline. The pipeline at pivotal time T corresponding to an optimal basic
feasible solution X = {xij } of (P2 ) is given by p =
xij .

{(i,j)/tij =T }

Construct related problem (RP − T ) as:
c∗ij xij

min

X∈S

where

(RP-T)

i∈I j∈J



0
∗
cij = 1, if tij = T,


∞

if tij < T,
if tij > T.

Remark 2.2. The optimal value of the problem (RP-T) yields minimum pipeline
at time T .
So, mathematically the problem can be formulated as

min (Z, p),

X∈S

(2.1)

where Z is the minimum cost of transportation problem and p is the minimum
pipeline at time T , yielded by optimal solution of (P2 ) and (RP-T), respectively.
Pair. A cost pipeline pair at given time of transportation T is denoted by (T :
Z, p).
Dominated pair. A pair (T : Z, p) is called dominated pair at pivotal time T
if there exists a pair (T : Z ∗ , p∗ ) such (Z, p) ≥ (Z ∗ , p∗ ) i.e. Z ≥ Z ∗ and p ≥ p∗
with strict inequality holding least at one place.
Non-dominated pair. A pair which is not dominated is called a non-dominated
pair.
Efficient point. A solution yielding a non-dominated pair is called an efficient
point.


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3. NOTATIONS
qth Efficient pair (T : Zq , pq ), (q ≥ 2). The q th efficient pair is that
member of Lq−1 for which Zq = min{Z | (T : Z, p) ∈ Lq−1 }, where the set Lq for
q ≥ 1 is defined below.
For q ≥ 1, following notations are introduced.
X q is the set of all basic feasible solutions yielding the q th efficient pair ={X qh | h =
1 to sq }.

B qh is the set of basic cells of solution X qh .
∆ij is the relative cost co-efficient for a basic feasible solution of problem (P2 ) for
a cell (i, j).
∆ij is the relative cost co-efficient for a basic feasible solution of problem (RP-T)
for the cell (i, j).
ˆ qh is the basic feasible solution derived from X qh by a single pivot operation
X
such that the corresponding time of transportation remains T, that is
max

{(i,j) | x
ˆqh
ij >0}

tij = T.

For each h = 1, 2 . . . sq , define


qh
qh
∆ij < 0, ∆ij > 0, x
ˆqh
ˆqh


ij = xlm , x
lm = 0, (l, m) ∈ B
N qh = (i, j) ∈
/ B qh and

max
trw = T


qh
{(r,w) | x
ˆrw >0}

Collecting those nonbasic cells, so that the entry of which into the current basis
corresponds to a q th efficient pair, increases the cost of transportation and reduces
the pipeline.
qh
qh
qh
Dqh = {(T : Z, p)|Z = Zq − ∆ij xqh
lm , p = pq − ∆ij xlm , (l, m) ∈ B , (i, j) ∈ N },
is the collection of all the pairs (T : Z, p), where Z is the increased cost and p
is the reduced pipelineobtained by entering nonbasic cells from the set N qh in a
single pivot operation.
sq

Dq =

Dqh
h=1

Lq = Lq−1 − {(T : Zq , pq )} for q ≥ 2, where L1 = ∅ and the list Lq is given by:
Lq = Lq ∪Dq −{(T : Z, p) | (T : Z, p) is a dominated pair in Lq ∪Dq }
E is the set of efficient pairs (T : Zi , pi ), i = 1 to M.

4. THEORETICAL DEVELOPMENT
This section discusses the main theoretical results, which lead to the convergence of procedure given in Section-4.


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Theorem 4.3. There exists a basic feasible solution yielding the first efficient pair
(T : Z1 , p1 ).
Proof : Let X0 be an optimal basic feasible solution (OBFS) of problem (P2 )
giving Z1 as the minimum cost at pivotal time T . Let B0 be the set of basic cells
of the solution X0 . Construct set
N0 = {(i, j) ∈
/ B0 | ∆ij = 0, ∆ij > 0}
If N0 = ∅, then X0 gives the first efficient pair (T : Z1 , p1 ), else choose (s, t) ∈ N0
such ∆st = max{∆ij | (i, j) ∈ N0 } and enter cell (s, t) into basis B0 . This will
reduce the pipeline at the same cost Z1 . Let the basic feasible solution thus
obtained be X1 with basis B1 . Let N1 = {(i, j) ∈
/ B1 | ∆ij = 0, ∆ij > 0}. Again,
if N1 = ∅, X1 gives the pair (T : Z1 , p1 ) and when N1 = ∅, entry of cell (d, e) ∈ N1 ,
into basis B1 , where ∆de = max{∆ij | (i, j) ∈ N1 } further decreases the pipeline
at cost Z1 . Continuing likewise, a sequence of solutions is constructed till a stage
is reached at which Ng = ∅.
Denote Xg by X 11 . Thus, X 11 is a basic feasible solution with basis B 11 , giving
pair (T : Z1 , p1 ).
Remark 4.4. If set H = {(i, j) ∈
/ B 11 | ∆ij = 0, ∆ij = 0}, then X 1 is the set of
all basic feasible solutions, each obtained as a result of entering a cell of H into
basis B 11 .

Corollary 4.5. Every efficient pair (T : Zq , pq ), q ≥ 2, is attainable at a basic
feasible solution.
Proof : The second efficient pair (T : Z2 , p2 ) is that member of L1 for which
Z2 = min{Z | (T : Z, p) ∈ L1 }. Thus (T : Z2 , p2 ) is attained at a basic feasible
solution. By definition of Dq , Lq (q ≥ 2), every pair (T : Zq , pq ), q ≥ 3 also
corresponds to a basic feasible solution.
Remark 4.6. Corollary 4.5 justifies that E is a finite set.
Theorem 4.7. For any efficient pair (T : Zq , pq ), pq is the minimum pipeline at
cost Zq and Zq is the minimum cost at pipeline pq .
Proof : It is sufficient to show for each h = 1 to sq , the sets


qh
qh
∆ij = 0, ∆ij > 0, x
ˆqh
ˆqh


ij = xlm , x
lm = 0, (l, m) ∈ B
S1h = (i, j) ∈
/ B qh and
max
trw = T


qh
{(r,w) | x
ˆrw >0}

and


S2h =




qh
qh
∆ij > 0, ∆ij = 0, x
ˆqh
ˆqh

ij = xlm , x
lm = 0, (l, m) ∈ B

(i, j) ∈
/ B qh and

max

{(r,w) | x
ˆqh
rw >0}

trw = T



Vikas Sharma, Rita Malhotra, Vanita Verma / A Cost And Pipeline

203

are both empty. Suppose, on the contrary, that S1h = ∅ for some h, 1 ≤ h ≤
sq . Then there exists a cell (i, j) in S1h such that ∆ij = 0, ∆ij > 0, x
ˆqh
ij =
qh
qh
qh
xlm , (l, m) ∈ B , x
ˆlm = 0. The entry of such a cell (i, j) into the basis of the
solution X qh will result in a pair (T : Zq , p) with p < pq . This contradicts the
non-dominance of (T : Zq , pq ). Hence S1h = ∅. Similarly S2h = ∅ ∀ h = 1 to sq .
Corollary 4.8. If (T : Z, p) is a pair and Z = Z , p = p where (T : Z , p ) is
an efficient pair, then (T : Z, p) is dominated.
Theorem 4.9. A non-dominated pair (T : Z, p) not in E satisfies the relation
M

Z=

M

λi Z i , p ≤
i=1

M

λi pi ,
i=1

λi = 1, λi ≥ 0, i = 1 to M.
i=1

Proof : Since (T : Z, p) is a non-dominated pair not in E, therefore (T : Z, p) ∈
/
M

Li and (T : Z, p) does not correspond to a basic feasible solution. Thus (T :
i=1

Z, p) is yielded by a feasible solution. Clearly Z1 < Z < ZM . There exist scalars λi
M

not all zero such that Z =

M

i=1

M

λi = 1, λi ≥ 0, i = 1 to M. Let p∗ =

λi Z i ,
i=1

λi pi .
i=1

M

Since (T : Z, p) is non-dominated, p ≤ p∗ =

λi pi .
i=1

Theorem 4.10. If (T : Z, p) is a pair with Z = Zi , i = 1 to M and Z1 < Z < ZM
then there exists p ≤ p such that (T : Z, p ) is a non-dominated pair.
Proof : Since Z = Zi , i = 1 to M, therefore (T : Z, p) ∈
/ E.
Let Zk < Z < Zk+1 , k ∈ {1, 2, . . . M }. Then there exists λ such that Z =
λZk + (1 − λ)Zk+1 , 0 < λ < 1. Consider p∗ = λpk + (1 − λ)pk+1 . Clearly
pk+1 < p∗ < pk . Also p∗ ≤ p, because if p∗ > p then (T : Z, p∗ ) is dominated
by (T : Z, p) which is a contradiction as (T : Z, p∗ ) being a convex combination
of adjacent efficient pairs, must be non-dominated. Setting p∗ = p , the desired
result is obtained.
Theorem 4.11. E is the exhaustive set of efficient pairs.
Proof : The proof is divided into two pairs.
Case I. No efficient pair other than the ones in E can be derived from a dominated
pair.
Suppose on the contrary that (T : Z , p ) is an efficient pair derived from a
dominated pair (T : Z, p), such that (T : Z , p ) = (T : Zi , pi ), i = 1 to M.
Now (T : Z , p ) is a non-dominated pair not in E and so from Corollary 4.3, it
follows that it does not correspond to a basic feasible solution. This violates the
very character of an efficient pair.
Case II. Any pair (T : Z, p) derived from an efficient pair (T : Zq , pq ) with

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Z < Zq , p > pq is either identical with one of the efficient pairs (T : Zi , pi ), i =
1 to q − 1 or is a dominated pair or is a convex combination of efficient pairs
(T : Zi , pi ), i = 1 to M. Now Z1 being the global minimum cost at time T, Z1 ≤
Z < Zq . If p > p1 , then (T : Z, p) is dominated by (T : Z1 , p1 ) and the conclusion
follows. Let now p ≤ p1 . Thus pq < p ≤ p1 . Two exhaustive cases arise:
(i) Z = Zi . for some i ∈ {1, 2, . . . q − 1} In this case p = p1 because if p1 > p then
(T : Z1 , p1 ) is dominated by (T : Z1 , p) which contradicts the efficient character
of (T : Z1 , p1 ). Thus in this case (T : Z, p) is identical with (T : Z1 , p1 )
(ii) Z = Zi . Let Zr < Z < Zr+1 where r ≥ 1, q ≥ r+1. Let Z = λZr +(1−λ)Zr+1
where 0 < λ < 1 and p∗ = λpr + (1 − λ)pr+1 . Thus pr+1 < p∗ < pr . If p∗ > p,
then (T : Z, p∗ ) is dominated by (T : Z, p) which is a contradiction, because
(T : Z, p∗ ) is a convex combination of two adjacent efficient pairs and therefore
must be a non-dominated pair. Thus p∗ ≤ p, if p∗ < p then clearly (T : Z, p) is a
dominated pair and if p∗ = p, then (T : Z, p) is a convex combination of efficient
pairs in E.
Theorem 4.12. The last pair in E gives the global minimum pipeline at pivotal
time T.
Proof : Since (T : ZM , pM ) is the last pair, LM −1 is a singleton, viz. LM −1 =
{(T : ZM , pM )} and DM = ∅. Thus N M h = ∅. Therefore there does not exist
any cell (i, j) ∈
/ B M h with ∆ij < 0, ∆ij > 0, the entry of which into the basis of
Mh
solution X
results in a solution with time of transportation equal to T . Thus

the only possibility for cell (i, j) ∈
/ B M h with ∆ > 0 are ∆ij = 0, ∆ij > 0 or
∆ij > 0, ∆ij > 0. In both cases, the fact that (T : ZM , pM ) is efficient, is
contradicted. Hence ∆ij ≯ 0. Thus the set P given by


P =



(i, j) ∈
/ B M h , h = 1 to sM


h
Mh
h
Mh

∆ij > 0, x
ˆM
ˆM
ij = xlm , x
lm = 0, (l, m) ∈ B
and

max

M h >0}
{(r,w) | x

ˆrw

trw = T

is empty. Thus the pipeline cannot be reduced further at pivotal time T of transportation. Hence pM is the global minimum pipeline at pivotal time T .

5. ALGORITHM FOR FINDING COST AND PIPELINE
TRADEOFF PAIRS
Initially set r = 0 and l = 0. Let the partition of various time routes be
given by t0 > t1 > t2 · · · > tk , where tk = min{tij | i ∈ I, j ∈ J}
Step 0. Solve the problem (P2 ) at time T = tl and go to the next step.
Step 1. If every optimal basic feasible solution of problem (P2 ) yield time T , then
declare T as pivotal time and go to the next step, otherwise set l = l + 1 and go
to Step 0.
Step 2 (Finding Ist efficient pair; (T : Z1 , p1 )). Read optimal basic feasible




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205

solution of problem (P2 ) corresponding to time T with minimum cost as Z1 and
pipeline p1 and corresponding basis as Br .
Step (2.a) Construct Nr = {(i, j) ∈
/ Br | ∆ij = 0, ∆ij > 0}, if Nr = ∅, then go
to Step (2.c), otherwise go to Step (2.b).
Step (2.b) Choose (s, t) ∈ Nr such that ∆st = max{∆ij | (i, j) ∈ Nr } and enter
cell (s, t) into the basis Br , set r = r + 1 and obtain new basic feasible

solution as Xr with basis Br and go to Step (2.a).
Step (2.c) Record (T : Z1 , p1 ) as the first efficient pair and the corresponding
basic feasible solution as X 11 with basis B 11 . Construct the set H = {(i, j) ∈
/
B 11 | ∆ij = 0, ∆ij = 0} (as mentioned in Remark 4.2), compute X 1h , h =
2, 3 . . . s1 and set X 1 = {X 1h , h = 1, 2 . . . s1 }, go to the next step.
Step 3. Initially record E = {(T : Z1 , p1 )}, set q=1 and go to the next step.
Step 4. Construct the set N qh , h = 1, 2 . . . sq . If N qh = ∅ for all h = 1, 2 . . . sq ,
go to terminal step, otherwise go to Step 5.
Step 5. Construct Dq , Lq and Lq . Then note the (q + 1)th efficient pair
given as (T : zq+1 , pq+1 ), where zq+1 = min{Z | (T : Z, p) ∈ Lq } and set
E = E ∪ {(T : Zq+1 , pq+1 )}, set q = q + 1 and go to Step 4.
Step 6 (Terminal step). Set E is the exhaustive set of efficient pairs corresponding to the pivotal time T ( as proved in Theorem 4.9).
Theorem 5.13. The algorithm terminates in finite number of steps.
Proof. By virtue of Corollory 4.5, it follows that each efficient pair at
pivotal time T corresponds to an extreme point of the set of feasible solutions of
(P2 ). As there is a finite number of extreme points of the feasible set, and the
choice of sets Lq and Lq is such that none of the extreme points is repeatedly
examined, the proposed algorithm terminates in finite number of steps.

6. NUMERICAL ILLUSTRATION

Consider a 4 × 6 transportation problem given by Table 1. The upper left
corner in each cell gives the time of transportation on the corresponding route and
the lower right corner gives the per unit cost on that route.


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Table 1
ai ↓
12

13

34

40
17

19
7

32
18

8

12

36

7
27

70
11

30
20

40
45

60
21
28

40
bj →

8

10

20

18

22

25

19

The partition of various time routes is given by t0 (= 45) > t1 (= 40) > t2 (=
36) > t3 (= 34) > t4 (= 21) > t5 (= 20) > t6 (= 18) > t7 (= 13) > t8 (= 12) > t9 (=
11) > t10 (= 7). Step 0. Solve the problem (P2 ) at time T = t0 (= 45) and go to

Step 1.
Step 1. An optimal feasible solution is depicted in the follwoing table.
Table 2
ai ↓
12

13

34

10

40
7

19
7

32
18

36
9

70
11

17
8

12
7

18
30

20

27
40

45

60
21

9
40
bj →

10

19
8

18

20
25

28
22

19

Since the (OFS) of this problem does not yield time T (= 45), therefore T = 45 is
not pivotal time. Set l = 1 and go to Step 0
Step 0. Solve the problem (P2 ) at time T = t1 (= 40) and go to Step 1.
Step 1. This problem has two alternate optimal feasible solutions and are depicted


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207

in Tables 3 and 4 below.

Table 3
ai ↓
12

13

34

10

40
7

19
7

32
18

36
9

70
11

17
8

12
7

18
30

20

27
40

M

60
21

9
40
bj →

19
8

10

20

18

25

28
22

19

Table 4
ai ↓
12

13

34

40

10

7
19

7

32
18

8
36

2
70
11

7
25

30
20

27
40

M

60

21

16
40
bj →

10

12
8

18

17
12

20
25

28
22

19

From Tables 3 and 4, it is observed that (OFS) of problem (P2 ) at time T = 40
yield the time of transportation as 36 and 40, respectively. Therefore, T = 40 is
not a pivotal time. Set l = 2 and go to Step 0.
Step 0. Solve the problem (P2 ) at time T = t2 (= 36) and go to Step 1.

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Step 1. An optimal feasible solution of this problem is depicted in Table 5.
Table 5
ai ↓
12

13

34

10

M
7

19
7

32
18

12

36
9

70

11

17
8
7

18
30

20

27
40

M

60
21

9
40
bj →

10

19
8

18

20
25

28
22

19

Since there is no alternate optimal solution to the one depicted in Table 5, T = 36
is a pivotal time. Go to Step 2.
Step 2. Note the pair (T : Z, p) = (36 : 1726, 18) and the corresponding basis
as B0 and go to Step (2.a).
Step (2.a). Corresponding to the basis B0 , depicted in Table 5, the value of ∆ij
and ∆ij for (i, j) ∈
/ B0 is calculated bz using the formula ∆ij = ui + vj − cij
and ∆ij = ui + vj − cij respectively, and is shown in the table below. Initially
u1 = u1 = 0.
(i,j)
∆ij
∆ij

(1,2)
-1
-34

(1,4)
-M-1
0

Table 6

(2,1)
1
-19

(2,4)
0
-16

(3,1)
1
-11

(3,3)
1-m
-2

Since the set N0 = {(i, j) ∈
/ B0 | ∆ij = 0, ∆ij > 0} = ∅, the current recorded pair
(36 : 1726, 18) is efficient, go to Step (2.c)
Step (2.c). Record the first efficient pair as (T : Z, p) = (36 : 1726, 18) with
11
11
11
the corresponding solution as X 1 = X 11 = {x11
11 = 10, x13 = 7, x22 = 9, x23 =
11
11
18, x32 = 9, x34 = 19} and go to Step 3.
Step 3. Record E = {(36 : 1726, 18)}, set q = 1 and go to Step 4.
Step 4. Construct the set N 11 = {(2, 1), (3, 1)} from Table 6 and go to Step 5.

Step 5. Construct the set D1 (= D11 ) = {(36 : 1825, 9), (36 : 1916, 8)}, by entering
nonbasic cells from the set N 11 successively in single pivot operation. since there
are no dominated pairs in D1 , we have L1 = D1 and
(T : Z2 , p2 ) = min{Z | (T : Z, p) ∈ L1 },
we get (T : Z2 , p2 ) = (36 : 1825, 9). Update the set E = {(36 : 1726, 18), (36 :
1825, 9)}, set q = 2 and go to Step 4.
Step 4. The table depicting second efficient pair (T : Z2 , p2 ) = (36 : 1825, 9) is


Vikas Sharma, Rita Malhotra, Vanita Verma / A Cost And Pipeline

209

shown below
Table 7
ai ↓
12

13

34

1

M
16

19
7

32
18

36
18

70
11

17
8

12
7

9
30

20

27
40

M

60
21

9

19
40

bj →

8

10

20

18

28
22

25

19

Construct the set N 21 = {(2, 1), (2, 4)} on the same lines as constructed above the
set N0 . Since N 21 = ∅, go to Step 5.
Step 5. Constructing the set D2 = {(36 : 1844, 8), (36 : 1852, 8)}, we have
L2 = L1 \ {(36 : 1825, 9)} = (36 : 1916, 8)
and L2 = {(36 : 1844, 8)}. Therefore, the third efficient pair is given as (36 :
1844, 8), update the set E = {(36 : 1726, 18), (36 : 1825, 9), (36 : 1844, 8)}. Set
q = 3 and go to Step 4.
Step 4. Table depicting the third efficient pair is given below
Table 8
ai ↓

12

13

34

M
17

19
7

32
18

1

36
18

70
11

17
8

12
7

8

30

20

27
40

M

60
21

9

19
40

bj →

10

8
18

20
25

28
22

19

Since the set N 3h = ∅, go to Step 6.
Step 6. The exhaustive set of efficient pairs corresponding to the pivotal time
T = 36 is given by E = {(36 : 1726, 18), (36 : 1825, 9), (36 : 1844, 8)}


210

Vikas Sharma, Rita Malhotra, Vanita Verma / A Cost And Pipeline

7. CONCLUDING REMARKS
1. In this paper, we have presented an algorithm, which enumerates all the
independent, non-dominated cost-pipeline pairs called efficient pairs, which
correspond to basic feasible solutions (BFS) starting from the minimum cost
solution at pivotal time T chosen. When the time taken for transportation is
not pivotal, there always exists an alternate solution of problem (P2 ), which
yields zero unit of pipeline at time T, thereby reducing the total time of
transportation from T to time T < T and yielding a unique cost pipeline
pair with zero pipeline; hence, there is no need to provide nondominated
solution.
2. It may be noted that in order to find all efficient pairs, cost minimization
transportation problems (P2 ) and RP-T are being solved repeatedly and by
corollary 4.3, each such pair corresponds to an extreme point of S. Therefore,
only finite number of such problems are to be solved. The best polynomial
running time for cost minimization transportation problem is o(mlogn(m +
nlogn)), where |I| = m, |J| = n (Orlin [6]). Hence, the proposed algorithm
is a polynomial time algorithm.
3. This problem can further be explored in the case when decision variable are
taken as bounded.

4. The problem of finding cost pipeline efficient pairs with positive pipelines at
a given particular time which may not be a pivotal time, can be explored
further.
Acknowledgement: Authors are thankful to University Grants Commission for
the financial assistance provided for this work and also grateful to Dr. Kalpana
Dahiya for the fruitful discussions.

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